Design of a Truss
Design of a Truss: In the following, we shall discuss three methods for designing a statically and kinematically determinate plane truss.
Method 1:We start with a single bar and add two members to form a triangle (see Fig.1). This basic element represents a rigid body. It may be extended by successively adding two members at a time and connecting them in such a way that the structure remains rigid (one has to avoid having two members along a straight line; such an improper arrangement is indicated by the dashed line in Fig. 1). A truss designed from a basic triangle as described is called a simple plane truss.
As can be verified by inspection, the relation
2 j = m 3..............................eq..(1)
is satisfied by the trusses shown in Fig. 1. For every additional joint in a simple truss, there are two additional members. Therefore Equation (1) remains valid. If the truss is supported insuch a way that there are r = 3 unknown reactions that completely constrain the truss, then it is statically determinate (see the
necessary condition 2 j = m r .).
Method 2: Two simple trusses are connected by three members (Fig. 1a). To ensure the rigidity of the system, the axes of the members must not be parallel or concurrent. The two simple trusses may also be connected by a joint and one member: the system in Fig. 6.4b has been obtained by replacing members 2 and 3 in Fig. 1a by the joint I. If two simple trusses are connected through one joint only, the system is nonrigid. To obtain a statically and kinematically determinate system, an additional support must be introduced; an example is given in Fig. 1c. This system represents a three-hinged arch. It should be noted that instead of connecting the two simple
trusses by one joint the trusses may be connected by two members that are not parallel and not concurrent. It can easily be verified that for the three examples shown in Fig. 2, the necessary condition 2 j = m r . is satisfied. Since these systems are completely constrained against motion, they are statically determinate.
Method 3: Consider a truss that has been designed according to the first or the second method. If we remove one member of the truss, the truss will become nonrigid. Therefore, we have to add one member at a different position in such a way that the truss will be rigid again. Since by doing so neither the number of the members nor the number of the joints is changed, condition 2 j = m r . is still satisfied.
An example of Method 3 is shown in Fig. 1. If we remove member 1 from the simple truss in Fig. 1a, the system is only partially constrained. Adding member 1′ yields the statically determinate truss shown in Fig. 1b.